ADVANCED COMPUTATIONAL MODELS AND ALGORITHMS Instructor: Dr. Gautam Das Lecture 3 January 29, 2009 Class notes by Alexandra Stefan

Review (see previous lecture): optimization problem, decision problem, verification problem; P, NP; polynomial time reduction; Overview of lecture: New definitions: Decision problem as a set, complement of a decision problem, Co-NP, NP-Complete Example problems used or mentioned: HCP, TSP, SP, PRIME, SATISFIABILITY (HCP = Hamiltonian Cycle, TSP = Traveling Salesman, SP = Shortest Path ) Other things to remember: Pay attention to the proper formulation of what constitutes the input of each problem. It is not always easy to produce a certificate for a decision problem.

Definitions NP Decision problems as sets: set of problems whose verification version can be solved in polynomial time. Imporatnt: You need to have a certificate Decision problems as sets: A decision problem takes an input and produces an ‘Yes’ or ‘No’ output. Any decision problem, X, can be defined as the set of inputs for which the answer is ‘Yes’. The complement of a decision problem: Given a decision problem X, the complement of X is the set of inputs for which the answer is ‘No’. Co-NP = set of decision problems whose complement problem is in NP. NP-Complete: A problem X is NP-Complete if: X is NP Any NP problem Y can be reduced in polynomial time to X. (“X should be the hardest problem in it’s class”) (Intuition: finding an NP-Complete problem is like finding the largest number in a set of numbers)

NP-Complete NP NP Co-NP P P

SP – Shortest Path SP SP is in NP: SP is in Co-NP: Set version of SP: Input: G (graph) ,s (vertex), t (vertex), x (value) Output: ‘Yes’ if there exists a path from s to t of length less or equal to x. ‘No’, otherwise. SP is in NP: Certificate: list of vertices Easy to check if the certificate is a valid path, to compute it’s length and compare it with x. SP is in Co-NP: Certificate: nothing use Dijkstra to get the shortest path and it’s length, y. This runs in polynomial time. Compare x with y. Set version of SP: The set of tuples: < G (graph) ,s (vertex), t (vertex), x (value) > for which the answer is ‘Yes’.

Hamiltonian Cycle Problem (HCP) Definition: Input: unweighted graph Output: Yes, if there exists a cycle that visits all vertices exactly once No, otherwise Examples Graph: Does it have a Hamiltonian cycle? Answer: No Yes

Hamiltonian Cycle Problem (HCP) Is HCP in NP ? Yes A certificate is a list of vertices It is easy to verify that the certificate constitutes a valid cycle in the graph and that it covers all vertices Is HCP in Co-NP ? yes Since HCP is in NP

NP-Complete How do you show that a problem is NP-Complete? You have to show that any problem in NP can be reduced to it. First problem that was shown to be NP-Complete is the SATISFIABILITY problem. It is known as Cook’s Theorem. Once we have an NP-Complete problem, X, to show that another NP problem, Y, is NP-Complete, we only need to show that X reduces in polynomial time to Y.

NP-Complete Y X If X is NP-Complete and X reduces in polynomial time to Y then any problem can be reduced in polynomial time to Y by First reducing it to X and then reducing X to Y. (Here we used the property of polynomials of being closed under multiplication.)

(using reduction of X to Y) Polynomial reduction 3 equivalent ways of saying that X reduces in polynomial time to Y: reducer Solver of Y Solver of X (using reduction of X to Y) Input for X answer Set mapping after reduction X Y Universe of inputs for X Universe of inputs for Y X Y

PRIME NP Co-NP P PRIME First believed to be NP. Input: number n (of m bits) Output: Is n a prime number or not? First believed to be NP. First shown to be both in NP and Co-NP. Then shown to be in P. NP Co-NP P PRIMES PRIMES